3.55 \(\int x \tan ^{-1}(x) \, dx\)

Optimal. Leaf size=21 \[ \frac{1}{2} x^2 \tan ^{-1}(x)-\frac{x}{2}+\frac{1}{2} \tan ^{-1}(x) \]

[Out]

-x/2 + ArcTan[x]/2 + (x^2*ArcTan[x])/2

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Rubi [A]  time = 0.0085168, antiderivative size = 21, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 4, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.75, Rules used = {4852, 321, 203} \[ \frac{1}{2} x^2 \tan ^{-1}(x)-\frac{x}{2}+\frac{1}{2} \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[x*ArcTan[x],x]

[Out]

-x/2 + ArcTan[x]/2 + (x^2*ArcTan[x])/2

Rule 4852

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTa
n[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTan[c*x])^(p - 1))/(1 + c^
2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int x \tan ^{-1}(x) \, dx &=\frac{1}{2} x^2 \tan ^{-1}(x)-\frac{1}{2} \int \frac{x^2}{1+x^2} \, dx\\ &=-\frac{x}{2}+\frac{1}{2} x^2 \tan ^{-1}(x)+\frac{1}{2} \int \frac{1}{1+x^2} \, dx\\ &=-\frac{x}{2}+\frac{1}{2} \tan ^{-1}(x)+\frac{1}{2} x^2 \tan ^{-1}(x)\\ \end{align*}

Mathematica [A]  time = 0.0025206, size = 21, normalized size = 1. \[ \frac{1}{2} x^2 \tan ^{-1}(x)-\frac{x}{2}+\frac{1}{2} \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcTan[x],x]

[Out]

-x/2 + ArcTan[x]/2 + (x^2*ArcTan[x])/2

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Maple [A]  time = 0., size = 16, normalized size = 0.8 \begin{align*} -{\frac{x}{2}}+{\frac{\arctan \left ( x \right ) }{2}}+{\frac{{x}^{2}\arctan \left ( x \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arctan(x),x)

[Out]

-1/2*x+1/2*arctan(x)+1/2*x^2*arctan(x)

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Maxima [A]  time = 1.41976, size = 20, normalized size = 0.95 \begin{align*} \frac{1}{2} \, x^{2} \arctan \left (x\right ) - \frac{1}{2} \, x + \frac{1}{2} \, \arctan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctan(x),x, algorithm="maxima")

[Out]

1/2*x^2*arctan(x) - 1/2*x + 1/2*arctan(x)

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Fricas [A]  time = 2.2782, size = 45, normalized size = 2.14 \begin{align*} \frac{1}{2} \,{\left (x^{2} + 1\right )} \arctan \left (x\right ) - \frac{1}{2} \, x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctan(x),x, algorithm="fricas")

[Out]

1/2*(x^2 + 1)*arctan(x) - 1/2*x

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Sympy [A]  time = 0.333704, size = 15, normalized size = 0.71 \begin{align*} \frac{x^{2} \operatorname{atan}{\left (x \right )}}{2} - \frac{x}{2} + \frac{\operatorname{atan}{\left (x \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*atan(x),x)

[Out]

x**2*atan(x)/2 - x/2 + atan(x)/2

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Giac [A]  time = 1.05583, size = 20, normalized size = 0.95 \begin{align*} \frac{1}{2} \, x^{2} \arctan \left (x\right ) - \frac{1}{2} \, x + \frac{1}{2} \, \arctan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctan(x),x, algorithm="giac")

[Out]

1/2*x^2*arctan(x) - 1/2*x + 1/2*arctan(x)